Well, for example to stress the extremely low importance of what letter we use to denote some variable/unknown , something some students have big trouble with sometimes.
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DonAntonioAug 15 '12 at 13:11

Maybe he is reffering to $r$ as a real number. In which case this inequality also holds. It holds for $r \geq 1$ and $r \leq 0$.
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clarkAug 15 '12 at 13:18

I kinda see variables as if they were placeholders, like $\diamond^\diamond$ even if they are $x^y$, $z^g$, $p^o$, etc. I just thought this switch was some standard mathematical method intended to mean something else.
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VoyskaAug 15 '12 at 13:20

@clark Here it is used in an argument where $n$ and $r$ both represent only integers. That is also another common reason, though.
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process91Aug 15 '12 at 13:31

2 Answers
2

Courant & Robbins' style in that book was to reserve the letter $n$ in an inductive argument for the general case, like this:

We want to prove that some statement $P(n)$ is true for all natural numbers $n$.

First we prove that it is true for $P(0)$.

Now we assume it is true for some natural number $r$, and we will prove it must also be true for $r+1$. That is, we will show $P(r)\implies P(r+1)$.

Here the authors are trying to make explicit the idea that $n$ represents any natural number, and in their induction step they are choosing one particular natural number $r$. This is simply the way that induction is introduced in the book at the top of page 11.

The essential idea in the preceding arguments is to establish a general theorem $A$ for all values of $n$ by successively proving a sequence of special cases, $A_1, A_2, \dots$ . The possibility of doing this depends on two things: a) There is a general method for showing that if any statement $A_r$ is true then the next statement, $A_{r+1}$, will also be true. b) The first statement $A_1$ is known to be true.

Once you are comfortable with this idea, many people simply use $n$ in the second step as well, remembering that in that second step $n$ represents a particular natural number.

The authors are proving this by induction. They switch from $n$ to $r$ to help readers who would be confused if they assumed something for $n$ when they're trying to prove it in the first place. So they assume the result for $r$ and show how to extend it to $r+1$, thus proving it for all $n$. No big deal.